EXAMPLE 1 Homogeneous Linear ODEs: Superposition of Solutions

The functions y = cos x and y = sin x are solutions of the homogeneous linear ODE

for all x. We verify this by differentiation and substitution. We obtain (cos x)″ = −cos x hence

Similarly for y = sin x (verify!). We can go an important step further. We multiply cos x by any constant, for instance, 4.7, and sin x by, say, −2, and take the sum of the results, claiming that it is a solution. Indeed, differentiation and substitution gives

THEOREM 1 Fundamental Theorem for the Homogeneous Linear ODE (2)

For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

PROOF

Let y₁ and y₂ be solutions of (2) on I. Then by substituting y = c₁y₁ + c₂y₂ and its derivatives into (2), and using the familiar rule , etc., we get

EXAMPLE 2 A Nonhomogeneous Linear ODE

Verify by substitution that the functions y = 1 + cos x and y = 1 + sin x are solutions of the nonhomogeneous linear ODE

but their sum is not a solution. Neither is, for instance, 2(1 + cos x) or 5(1 + sin x).

EXAMPLE 3 A Nonlinear ODE

Verify by substitution that the functions y = x² and y = 1 are solutions of the nonlinear ODE

but their sum is not a solution. Neither is −x², so you cannot even multiply by −1!

EXAMPLE 4 Initial Value Problem

Solve the initial value problem

Solution. Step 1. General solution. The functions cos x and sin x are solutions of the ODE (by Example 1), and we take

Fig. 29. Particular solution and initial tangent in Example 4

This will turn out to be a general solution as defined below.

Step 2. Particular solution. We need the derivative y′ = −c₁ sin x + c₂ cos x. From this and the initial values we obtain, since cos 0 = 1 and sin 0 = 0,

This gives as the solution of our initial value problem the particular solution

Figure 29 shows that at x = 0 it has the value 3.0 and the slope −0.5, so that its tangent intersects the x-axis at x = 3.0/0.5 = 6.0. (The scales on the axes differ!)

DEFINITION General Solution, Basis, Particular Solution

A general solution of an ODE (2) on an open interval I is a solution (5) in which and are solutions of (2) on I that are not proportional, and c₁ and c₂ are arbitrary constants. These y₁, y₂ are called a basis (or a fundamental system) of solutions of (2) on I.

A particular solution of (2) on I is obtained if we assign specific values to c₁ and c₂ in (5).

DEFINITION Basis (Reformulated)

A basis of solutions of (2) on an open interval I is a pair of linearly independent solutions of (2) on I.

EXAMPLE 5 Basis, General Solution, Particular Solution

cos x and sin x in Example 4 form a basis of solutions of the ODE y″ + y = 0 for all x because their quotient is cot x ≠ const (or tan x ≠ const). Hence y = c₁ cos x + c₂ sin x is a general solution. The solution y = 3.0 cos x − 0.5 sin x of the initial value problem is a particular solution.

EXAMPLE 6 Basis, General Solution, Particular Solution

Verify by substitution that y₁ = e^x and y₂ = e^−x are solutions of the ODE y″ − y = 0. Then solve the initial value problem

Solution. (e^x)″ − e^x = 0 and (e^−x)″ − e^−x = 0 show that e^x and e^−x are solutions. They are not proportional, e^x/e^−x = e^2x ≠ const. Hence e^x, e^−x form a basis for all x. We now write down the corresponding general solution and its derivative and equate their values at 0 to the given initial conditions,

By addition and subtraction, c₁ = 2, c₂ = 4, so that the answer is y = 2e^x + 4e^−x. This is the particular solution satisfying the two initial conditions.

EXAMPLE 7 Reduction of Order if a Solution Is Known. Basis

Find a basis of solutions of the ODE

Solution. Inspection shows that y₁ = x is a solution because y′₁ = 1 and y″₁ = 0, so that the first term vanishes identically and the second and third terms cancel. The idea of the method is to substitute

into the ODE. This gives

ux and −xu cancel and we are left with the following ODE, which we divide by x, order, and simplify,

This ODE is of first order in ν = u′, namely, (x² − x)ν′ + (x − 2)ν = 0. Separation of variables and integration gives

We need no constant of integration because we want to obtain a particular solution; similarly in the next integration. Taking exponents and integrating again, we obtain

Since y₁ = x and y₂ = x ln |x| + 1 are linearly independent (their quotient is not constant), we have obtained a basis of solutions, valid for all positive x.

PROBLEM SET 2.1

REDUCTION OF ORDER is important because it gives a simpler ODE. A general second-order ODE F(x, y, y′, y″) = 0, linear or not, can be reduced to first order if y does not occur explicitly (Prob. 1) or if x does not occur explicitly (Prob. 2) or if the ODE is homogeneous linear and we know a solution (see the text).

1. Reduction. Show that F(x, y′, y″) = 0 can be reduced to first order in z = y′ (from which y follows by integration). Give two examples of your own.
2. Reduction. Show that F(y, y′, y″) = 0 can be reduced to a first-order ODE with y as the independent variable and y″ = (dz/dy)z, where z = y′ derive this by the chain rule. Give two examples.

3–10 REDUCTION OF ORDER

Reduce to first order and solve, showing each step in detail.

• 3. y″ + y′ = 0
• 4. 2xy″ = 3y′
• 5. yy″ = 3y′²
• 6. xy″ + 2y′ + xy = 0, y₁ = (cos x)/x
• 7. y″ + y′³ sin y = 0
• 8. y″ = 1 + y′²
• 9. x²y″ − 5xy′ + 9y = 0, y₁ = x³
• 10. y″ + (1 + 1/y)y′² = 0

11–14 APPLICATIONS OF REDUCIBLE ODEs

• 11. Curve. Find the curve through the origin in the xy-plane which satisfies y″ = 2y′ and whose tangent at the origin has slope 1.
• 12. Hanging cable. It can be shown that the curve y(x) of an inextensible flexible homogeneous cable hanging between two fixed points is obtained by solving , where the constant k depends on the weight. This curve is called catenary (from Latin catena = the chain). Find and graph y(x), assuming that k = 1 and those fixed points are (−1, 0) and (1, 0) in a vertical xy-plane.
• 13. Motion. If, in the motion of a small body on a straight line, the sum of velocity and acceleration equals a positive constant, how will the distance y(t) depend on the initial velocity and position?
• 14. Motion. In a straight-line motion, let the velocity be the reciprocal of the acceleration. Find the distance y(t) for arbitrary initial position and velocity.

15–19 GENERAL SOLUTION. INITIAL VALUE PROBLEM (IVP)

(More in the next set.) (a) Verify that the given functions are linearly independent and form a basis of solutions of the given ODE. (b) Solve the IVP. Graph or sketch the solution.

• 15. 4y″ + 25y = 0, y(0) = 3.0, y′(0) = −2.5, cos 2.5x, sin 2.5x
• 16. y″ + 0.6y′ = 0.09y = 0, y(0) = 2.2, y′(0) = 0.14, e^−0.3x, xe^−0.3x
• 17. 4x²y″ − 3y = 0, y(1) = −3, y′(1) = 0, x^3/2, x^−1/2
• 18. x²y″ − xy′ + y = 0, y(1) = 4.3, y′(1) = 0.5, x, x ln x
• 19. y″ + 2y′ + 2y = 0, y(0) = 0, y′(0) = 15, e^−x cos x, e^−x sin x
• 20. CAS PROJECT. Linear Independence. Write a program for testing linear independence and dependence. Try it out on some of the problems in this and the next problem set and on examples of your own.

EXAMPLE 1 General Solution in the Case of Distinct Real Roots

We can now solve y″ − y = 0 in Example 6 of Sec. 2.1 systematically. The characteristic equation is λ² − 1 = 0. Its roots are λ₁ = 1 and λ = −1. Hence a basis of solutions is e^x and e^−x and gives the same general solution as before,

EXAMPLE 2 Initial Value Problem in the Case of Distinct Real Roots

Solve the initial value problem

Solution. Step 1. General solution. The characteristic equation is

Its roots are

so that we obtain the general solution

Step 2. Particular solution. Since y′ (x) = c₁e^x − 2c₂e^−2x, we obtain from the general solution and the initial conditions

Hence c₁ = 1 and c₂ = 3. This gives the answer y = e^x + 3e^−2x. Figure 30 shows that the curve begins at y = 4 with a negative slope (−5, but note that the axes have different scales!), in agreement with the initial conditions.

EXAMPLE 3 General Solution in the Case of a Double Root

The characteristic equation of the ODE y″ + 6y′ + 9y = 0 is λ² + 6λ + 9 = (λ + 3)² = 0. It has the double root λ = −3. Hence a basis is e^−3x and xe^−3x. The corresponding general solution is y = (c₁ + c₂x)e^−3x.

EXAMPLE 4 Initial Value Problem in the Case of a Double Root

Solve the initial value problem

Solution. The characteristic equation is λ² + λ + 0.25 = (λ + 0.5)² = 0. It has the double root λ = −0.5. This gives the general solution

We need its derivative

From this and the initial conditions we obtain

The particular solution of the initial value problem is y = (3 − 2x)e^−0.5x. See Fig. 31.

EXAMPLE 5 Complex Roots. Initial Value Problem

Solve the initial value problem

Solution. Step 1. General solution. The characteristic equation is λ² + 0.4λ + 9.04 = 0. It has the roots −0.2 ± 3i. Hence ω = 3, and a general solution (9) is

Step 2. Particular solution. The first initial condition gives y(0) = A = 0. The remaining expression is y = Be^−0.2x sin 3x. We need the derivative (chain rule!)

From this and the second initial condition we obtain y′(0) = 3B = 3. Hence B = 1. Our solution is

Figure 32 shows y and the curves of e^−0.2x and −e^−0.2x (dashed), between which the curve of y oscillates. Such “damped vibrations” (with x = t being time) have important mechanical and electrical applications, as we shall soon see (in Sec. 2.4).

EXAMPLE 6 Complex Roots

A general solution of the ODE

is

With ω = 1 this confirms Example 4 in Sec. 2.1.

PROBLEM SET 2.2

1–15 GENERAL SOLUTION

Find a general solution. Check your answer by substitution. ODEs of this kind have important applications to be discussed in Secs. 2.4, 2.7, and 2.9.

1. 4y″ − 25y = 0
2. y″ + 36y = 0
3. y″ + 6y′ + 8.96y = 0
4. y″ + 4y′ + (π² + 4)y = 0
5. y″ + 2πy′ + π²y = 0
6. 10y″ − 32y′ + 25.6y = 0
7. y″ + 4.5y′ = 0
8. y″ + y′ + 3.25y = 0
9. y″ + 1.8y′ − 2.08y = 0
10. 100y″ + 240y′ + (196π² + 144)y = 0
11. 4y″ − 4y′ − 3y = 0
12. y″ + 9y′ + 20y = 0
13. 9y″ − 30y′ + 25y = 0
14. y″ + 2k²y′ + k⁴y = 0
15. y″ + 0.54y′ + (0.0729 + π)y = 0

16–20 FIND AN ODE

y″ + ay′ + by = 0 for the given basis.

• 16. e^2.6x, e^−4.3x
• 17.
• 18. cos 2πx, sin 2πx
• 19. 3^(−2+i)x, e^(−2−i)x
• 20. e^−3.1x cos 2.1x, e^−3.1x sin 2.1 x

21–30 INITIAL VALUES PROBLEMS

Solve the IVP. Check that your answer satisfies the ODE as well as the initial conditions. Show the details of your work.

• 21. y″ + 25y = 0, y(0) = 4.6, y′(0) = −1.2
• 22. The ODE in Prob. 4,
• 23. y″ + y′ − 6y = 0, y(0) = 10, y′(0) = 0
• 24. 4y″ − 4y′ − 3y = 0, y(−2) = e, y′(−2) = −e/2
• 25. y″ − y = 0, y(0) = 2, y′(0) = −2
• 26. y″ − k²y = 0 (k ≠ 0), y(0) = 1, y′(0) = 1
• 27. The ODE in Prob. 5,

  y(0) = 4.5, y′(0) = −4.5π − 1 = 13.137

• 28. 8y″ − 2y′ − y = 0, y(0) = − 0.2, y′(0) = −0.325
• 29. The ODE in Prob. 15, y(0) = 0, y′(0) = 1
• 30. 9y″ − 30y′ + 25y = 0, y(0) = 3.3, y′(0) = 10.0

31–36 LINEAR INDEPENDENCE is of basic importance, in this chapter, in connection with general solutions, as explained in the text. Are the following functions linearly independent on the given interval? Show the details of your work.

• 31. e^kx, xe^kx, any interval
• 32. e^ax, e^−ax, x > 0
• 33. x², x² ln x, x > 1
• 34. ln x, ln (x³), x > 1
• 35. sin 2x, cos x sin x, x < 0
• 36.
• 37. Instability. Solve y″ − y = 0 for the initial conditions y(0) = 1, y′(0) = −1. Then change the initial conditions to y(0) = 1.001, y′(0) = −0.999 and explain why this small change of 0.001 at t = 0 causes a large change later, e.g., 22 at t = 10. This is instability: a small initial difference in setting a quantity (a current, for instance) becomes larger and larger with time t. This is undesirable.
• 38. TEAM PROJECT. General Properties of Solutions (a) Coefficient formulas. Show how a and b in (1) can be expressed in terms of λ₁ and λ₂. Explain how these formulas can be used in constructing equations for given bases.

  (b) Root zero. Solve y″ + 4y′ = 0 (i) by the present method, and (ii) by reduction to first order. Can you explain why the result must be the same in both cases? Can you do the same for a general ODE y″ + ay′ = 0?

  (c) Double root. Verify directly that xe^λx with λ = −a/2 is a solution of (1) in the case of a double root. Verify and explain why y = e^−2x is a solution of y″ − y′ − 6y = 0 but xe^−2x is not.

  (d) Limits. Double roots should be limiting cases of distinct roots λ₁, λ₂ as, say, λ₂ → λ₁. Experiment with this idea. (Remember l'Hôpital's rule from calculus.) Can you arrive at ? Give it a try.

EXAMPLE 1 Factorization, Solution of an ODE

Factor P(D) = D² − 3D − 40I and solve P(D)y = 0.

Solution. D² − 3D − 40I = (D − 8I)(D + 5I) because I² = I. Now (D − 8I)y = y′ − 8y = 0 has the solution y₁ = e^8x. Similarly, the solution of (D + 5I)y = 0 is y₂ =e^−5x. This is a basis of P(D)y = 0 on any interval. From the factorization we obtain the ODE, as expected,

Verify that this agrees with the result of our method in Sec. 2.2. This is not unexpected because we factored in the same way as the characteristic polynomial P(λ) = λ² − 3λ − 40.

PROBLEM SET 2.3

1–5 APPLICATION OF DIFFERENTIAL OPERATORS

Apply the given operator to the given functions. Show all steps in detail.

1. D² + 2D; cosh 2x, e^−x + e^2x, cos x
2. D − 3I; 3x² + 3x, 3e^3x, cos 4x − sin 4x
3. (D − 2I)²; e^2x, xe^2x, e^−2x
4. D + 6I)²; 6x + sin 6x, xe^−6x
5. (D − 2I)(D + 3I); e^2x, xe^2x, e^−3x

6–12 GENERAL SOLUTION

Factor as in the text and solve.

• 6. (D² + 4.00D + 3.36I)y = 0
• 7. (4D² − I)y = 0
• 8. (D² + 3I)y = 0
• 9. (D² − 4.30D + 4.41I)y = 0
• 10. (D² + 4.80D + 5.76I)y = 0
• 11. (D² − 4.00D + 3.84I)y = 0
• 12. (D² + 3.0D + 2.51I)y = 0
• 13. Linear operator. Illustrate the linearity of L in (2) by taking c = 4, k = −6, y = e^2x, and w = cos 2x. Prove that L is linear.
• 14. Double root. If D² + aD + bI has distinct roots μ and λ, show that a particular solution is y = (e^μx − e^λx)/(μ − λ). Obtain from this a solution xe^λx by letting μ → λ and applying l'Hôpital's rule.
• 15. Definition of linearity. Show that the definition of linearity in the text is equivalent to the following. If L[y] and L[w] exist, then L[y + w] exists and L[cy] and L[kw] exist for all constants c and k, and L[y + w] = L[y] + L[w] as well as L[cy] = cL[y] and L[kw] = kL[w].

EXAMPLE 1 Harmonic Oscillation of an Undamped Mass–Spring System

If a mass–spring system with an iron ball of weight W = 98 nt (about 22 lb) can be regarded as undamped, and the spring is such that the ball stretches it 1.09 m (about 43 in.), how many cycles per minute will the system execute? What will its motion be if we pull the ball down from rest by 16 cm (about 6 in.) and let it start with zero initial velocity?

Solution. Hooke's law (1) with W as the force and 1.09 meter as the stretch gives W = 1.09k; thus k = W/1.09 = 98/1.09 = 90[kg/sec²] = 90 [nt>meter]. The mass is m = W/g = 98/9.8 = 10 [kg]. This gives the frequency .

From (4) and the initial conditions, y(0) = A = 0.16 [meter] and y′(0) = ω₀B = 0. Hence the motion is

If you have a chance of experimenting with a mass–spring system, don't miss it. You will be surprised about the good agreement between theory and experiment, usually within a fraction of one percent if you measure carefully.

EXAMPLE 2 The Three Cases of Damped Motion

How does the motion in Example 1 change if we change the damping constant c from one to another of the following three values, with y(0) = 0.16 and y′(0) = 0 as before?

Solution. It is interesting to see how the behavior of the system changes due to the effect of the damping, which takes energy from the system, so that the oscillations decrease in amplitude (Case III) or even disappear (Cases II and I).

(I) With m = 10 and k = 90, as in Example 1, the model is the initial value problem

The characteristic equation is 10λ² + 100λ + 90 = 10(λ + 9)(λ + 1) = 0. It has the roots −9 and −1. This gives the general solution

The initial conditions give c₁ + c₂ = 0.16, −9c₁ − c₂ = 0. The solution is c₁ = −0.02, c₂ = 0.18. Hence in the overdamped case the solution is

It approaches 0 as t → ∞. The approach is rapid; after a few seconds the solution is practically 0, that is, the iron ball is at rest.

(II) The model is as before, with c = 60 instead of 100. The characteristic equation now has the form 10λ² + 60λ + 90 = 10(λ + 3)² = 0. It has the double root −3. Hence the corresponding general solution is

The initial conditions give y(0) = c₁ = 0.16, y′(0) = c₂ − 3c₁ = 0, c₂ = 0.48. Hence in the critical case the solution is

It is always positive and decreases to 0 in a monotone fashion.

(III) The model now is 10y″ + 10y′ + 90y = 0. Since c = 10 is smaller than the critical c, we shall get oscillations. The characteristic equation is . It has the complex roots [see (4) in Sec. 2.2 with a = 1 and b = 9]

This gives the general solution

Thus y(0) = A = 0.16. We also need the derivative

Hence y′(0) = −0.5A + 2.96B = 0, B = 0.5A/2.96 = 0.027. This gives the solution

We see that these damped oscillations have a smaller frequency than the harmonic oscillations in Example 1 by about 1% (since 2.96 is smaller than 3.00 by about 1%). Their amplitude goes to zero. See Fig. 40.

PROBLEM SET 2.4

1–10 HARMONIC OSCILLATIONS (UNDAMPED MOTION)

1. Initial value problem. Find the harmonic motion (4) that starts from y₀ with initial velocity ν₀. Graph or sketch the solutions for ω₀ = π, y₀ = 1, and various ν₀ of your choice on common axes. At what t-values do all these curves intersect? Why?
2. Frequency. If a weight of 20 nt (about 4.5 lb) stretches a certain spring by 2 cm, what will the frequency of the corresponding harmonic oscillation be? The period?
3. Frequency. How does the frequency of the harmonic oscillation change if we (i) double the mass, (ii) take a spring of twice the modulus? First find qualitative answers by physics, then look at formulas.
4. Initial velocity. Could you make a harmonic oscillation move faster by giving the body a greater initial push?
5. Springs in parallel. What are the frequencies of vibration of a body of mass m = 5 kg (i) on a spring of modulus k₁ = 20 nt/m, (ii) on a spring of modulus k₂ = 45 nt/m, (iii) on the two springs in parallel? See Fig. 41.

   

   Fig. 41. Parallel springs (Problem 5)

6. Spring in series. If a body hangs on a spring s₁ of modulus k₁ = 8, which in turn hangs on a spring s₂ of modulus k₂ = 12, what is the modulus k of this combination of springs?
7. Pendulum. Find the frequency of oscillation of a pendulum of length L (Fig. 42), neglecting air resistance and the weight of the rod, and assuming θ to be so small that sin θ practically equals θ.

   

   Fig. 42. Pendulum (Problem 7)

8. Archimedian principle. This principle states that the buoyancy force equals the weight of the water displaced by the body (partly or totally submerged). The cylindrical buoy of diameter 60 cm in Fig. 43 is floating in water with its axis vertical. When depressed downward in the water and released, it vibrates with period 2 sec. What is its weight?

   

   Fig. 43. Buoy (Problem 8)

9. Vibration of water in a tube. If 1 liter of water (about 1.06 US quart) is vibrating up and down under the influence of gravitation in a U-shaped tube of diameter 2 cm (Fig. 44), what is the frequency? Neglect friction. First guess.

   

   Fig. 44. Tube (Problem 9)

10. TEAM PROJECT. Harmonic Motions of Similar Models. The unifying power of mathematical methods results to a large extent from the fact that different physical (or other) systems may have the same or very similar models. Illustrate this for the following three systems

    (a) Pendulum clock. A clock has a 1-meter pendulum. The clock ticks once for each time the pendulum completes a full swing, returning to its original position. How many times a minute does the clock tick?

    (b) Flat spring (Fig. 45). The harmonic oscillations of a flat spring with a body attached at one end and horizontally clamped at the other are also governed by (3). Find its motions, assuming that the body weighs 8 nt (about 1.8 lb), the system has its static equilibrium 1 cm below the horizontal line, and we let it start from this position with initial velocity 10 cm/sec.

    

    Fig. 45. Flat spring

    (c) Torsional vibrations (Fig. 46). Undamped torsional vibrations (rotations back and forth) of a wheel attached to an elastic thin rod or wire are governed by the equation I₀θ″ + Kθ = 0, where θ is the angle measured from the state of equilibrium. Solve this equation for K/I₀ = 13.69 sec⁻², initial angle 30° (= 0.5235 rad) and initial angular velocity 20° sec⁻¹ (= 0.349 rad · sec⁻¹).

    

    Fig. 46. Torsional vibrations

11–20 DAMPED MOTION

• 11. Overdamping. Show that for (7) to satisfy initial conditions y(0) = y₀ and ν(0) = ν₀ we must have c₁ = [(1 + α/β)y₀ + ν₀/β]/2 and c₂ = [(1 − α/β)y₀ − ν₀/β]/2.
• 12. Overdamping. Show that in the overdamped case, the body can pass through y = 0 at most once (Fig. 37).
• 13. Initial value problem. Find the critical motion (8) that starts from with initial velocity ν₀. Graph solution curves for α = 1, y₀ = 1 and several ν₀ such that (i) the curve does not intersect the t- axis, (ii) it intersects it at t = 1, 2, …, 5, respectively.
• 14. Shock absorber. What is the smallest value of the damping constant of a shock absorber in the suspension of a wheel of a car (consisting of a spring and an absorber) that will provide (theoretically) an oscillation-free ride if the mass of the car is 2000 kg and the spring constant equals 4500 kg/sec²?
• 15. Frequency. Find an approximation formula for ω* in terms of ω₀ by applying the binomial theorem in (9) and retaining only the first two terms. How good is the approximation in Example 2, III?
• 16. Maxima. Show that the maxima of an underdamped motion occur at equidistant t- values and find the distance.
• 17. Underdamping. Determine the values of t corresponding to the maxima and minima of the oscillation y(t) = e^−t sin t. Check your result by graphing y(t).
• 18. Logarithmic decrement. Show that the ratio of two consecutive maximum amplitudes of a damped oscillation (10) is constant, and the natural logarithm of this ratio called the logarithmic decrement, equals Δ = 2πα/ω*. Find Δ for the solutions of y″ + 2y′ + 5y = 0.
• 19. Damping constant. Consider an underdamped motion of a body of mass m = 0.5. If the time between two consecutive maxima is 3 sec and the maximum amplitude decreases to its initial value after 10 cycles, what is the damping constant of the system?
• 20. CAS PROJECT. Transition Between Cases I, II, III. Study this transition in terms of graphs of typical solutions. (Cf. Fig. 47.)

  (a) Avoiding unnecessary generality is part of good modeling. Show that the initial value problems (A) and (B),

  

  (B) the same with different c and y′(0) = −2 (instead of 0), will give practically as much information as a problem with other m, k, y(0), y′(0).

  (b) Consider (A). Choose suitable values of c, perhaps better ones than in Fig. 47, for the transition from Case III to II and I. Guess c for the curves in the figure.

  (c) Time to go to rest. Theoretically, this time is infinite (why?). Practically, the system is at rest when its motion has become very small, say, less than 0.1% of the initial displacement (this choice being up to us), that is in our case,

  

  In engineering constructions, damping can often be varied without too much trouble. Experimenting with your graphs, find empirically a relation between t₁ and c.

  (d) Solve (A) analytically. Give a reason why the solution c of y(t₂) = −0.001, with t₂ the solution of y′(t) = 0, will give you the best possible c satisfying (11).

  (e) Consider (B) empirically as in (a) and (b). What is the main difference between (B) and (A)?

  

  Fig. 47. CAS Project 20

EXAMPLE 1 General Solution in the Case of Different Real Roots

The Euler–Cauchy equation x²y″ + 1.5xy′ − 0.5y = 0 has the auxiliary equation m² + 0.5m − 0.5 = 0. The roots are 0.5 and −1. Hence a basis of solutions for all positive x is y₁ = x^0.5 and y₂ = 1/x and gives the general

EXAMPLE 2 General Solution in the Case of a Double Root

The Euler–Cauchy equation x²y″ − 5xy′ + 9y = 0 has the auxiliary equation m² − 6m + 9 = 0. It has the double root m = 3, so that a general solution for all positive x is

EXAMPLE 3 Real General Solution in the Case of Complex Roots

The Euler–Cauchy equation x²y″ + 0.6xy′ + 15.04y = 0 has the auxiliary equation m² − 0.4m + 16.04 = 0. The roots are complex conjugate, m₁ = 0.2 + 4i and m₂ = 0.2 − 4i, where . We now use the trick of writing x = e^lnx and obtain

Next we apply Euler's formula (11) in Sec. 2.2 with t = 4ln x to these two formulas. This gives

We now add these two formulas, so that the sine drops out, and divide the result by 2. Then we subtract the second formula from the first, so that the cosine drops out, and divide the result by 2i. This yields

respectively. By the superposition principle in Sec. 2.2 these are solutions of the Euler–Cauchy equation (1). Since their quotient cot(4 ln x) is not constant, they are linearly independent. Hence they form a basis of solutions, and the corresponding real general solution for all positive x is

Figure 48 shows typical solution curves in the three cases discussed, in particular the real basis functions in Examples 1 and 3.

EXAMPLE 4 Boundary Value Problem. Electric Potential Field Between Two Concentric Spheres

Find the electrostatic potential ν = ν(r) between two concentric spheres of radii r₁ = 5 cm and r₂ = 10 cm kept at potentials ν₁ = 110 V and ν₂ = 0, respectively.

Physical Information. v (r) is a solution of the Euler–Cauchy equation rν″ + 2ν′ = 0, where ν′ = dv/dr.

Solution. The auxiliary equation is m² + m = 0. It has the roots 0 and −1. This gives the general solution ν(r) = c₁ + c₂/r. From the “boundary conditions” (the potentials on the spheres) we obtain

By subtraction, c₂/10 = 110, c₂ = 1100. From the second equation, c₁ = −c₂/10 = −110. Answer: ν(r) = −110 + 1100/rV. Figure 49 shows that the potential is not a straight line, as it would be for a potential between two parallel plates. For example, on the sphere of radius 7.5 cm it is not 110/2 = 55 V, but considerably less. (What is it?)

PROBLEM SET 2.5

1. Double root. Verify directly by substitution that x^(1−a)/2 is a solution of (1) if (2) has a double root, but and are not solutions of (1) if the roots m₁ and m₂ of (2) are different.

2–11 GENERAL SOLUTION

Find a real general solution. Show the details of your work.

• 2. x²y″ − 20y = 0
• 3. 5x²y″ + 23xy′ 16.2y = 0
• 4. xy″ + 2y′ = 0
• 5. 4x²y″ + 5y = 0
• 6. x²y″ + 0.7xy′ − 0.1y = 0
• 7. (x²D² − 4xD + 6I)y = C
• 8. (x²D² − 3xD + 4I)y = 0
• 9. (x²D² − 0.2xD + 0.36I)y = 0
• 10. (x²D² − xD + 5I)y = 0
• 11. (x²D² − 3xD + 10I)y = 0

12–19 INITIAL VALUE PROBLEM

Solve and graph the solution. Show the details of your work.

• 12. x²y″ − 4xy′ + 6y = 0, y(1) = 0.4, y′(1) = 0
• 13. x²y″ + 3xy′ + 0.75y = 0, y(1) = 1, y′(1) = −1.5
• 14. x²y″ + xy′ + 9y = 0, y(1) = 0, y′(1) = 2.5
• 15. x²y″ + 3xy′ + y = 0, y(1) = 3.6, y′(1) = 0.4
• 16. (x²D² − 3xD + 4I)y = 0, y(1) = −π, y′(1) = 2π
• 17. (x²D² + xD + I)y = 0, y(1) = 1, y′(1) = 1
• 18. (9x²D² + 3xD + I)y = 0, y(1) = 1, y′(1) = 0
• 19. (x²D² − xD − 15I)y = 0, y(1) = 0.1, y′(1) = −4.5
• 20. TEAM PROJECT. Double Root

  (a) Derive a second linearly independent solution of (1) by reduction of order; but instead of using (9), Sec. 2.1, perform all steps directly for the present ODE (1).

  (b) Obtain x^m ln x by considering the solutions x^m and x^m+s of a suitable Euler–Cauchy equation and letting s → 0.

  (c) Verify by substitution that x^m ln x, m = (1 − a)/2, is a solution in the critical case.

  (d) Transform the Euler–Cauchy equation (1) into an ODE with constant coefficients by setting x = e^t(x > 0).

  (e) Obtain a second linearly independent solution of the Euler–Cauchy equation in the “critical case” from that of a constant-coefficient ODE.

THEOREM 1 Existence and Uniqueness Theorem for Initial Value Problems

If p(x) and q(x) are continuous functions on some open interval I (see Sec. 1.1) and x₀ is in I, then the initial value problem consisting of (1) and (2) has a unique solution y(x) on the interval I.

THEOREM 2 Linear Dependence and Independence of Solutions

Let the ODE (1) have continuous coefficients p(x) and q(x) on an open interval I. Then two solutions of y₁ and y₂ of (1) on I are linearly dependent on I if and only if their “Wronskian”

is 0 at some x₀ in I. Furthermore, if W = 0 at an x = x₀ in I, then W = 0 on I; hence, if there is an x₁ in I at which W is not 0, then y₁, y₂ are linearly independent on I.

PROOF

(a) Let y₁ and y₂ be linearly dependent on I. Then (5a) or (5b) holds on I. If (5a) holds, then

Similarly if (5b) holds.

(b) Conversely, we let W(y₁, y₂) = 0 for some x = x₀ and show that this implies linear dependence of y₁ and y₂ on I. We consider the linear system of equations in the unknowns k₁, k₂

To eliminate k₂, multiply the first equation by y′₂ and the second by −y₂ and add the resulting equations. This gives

Similarly, to eliminate k₁, multiply the first equation by −y′₁ and the second by y₁ and add the resulting equations. This gives

If W were not 0 at x₀, we could divide by W and conclude that k₁ = k₂ = 0. Since W is 0, division is not possible, and the system has a solution for which k₁ and k₂ are not both 0. Using these numbers k₁, k₂, we introduce the function

Since (1) is homogeneous linear, Fundamental Theorem 1 in Sec. 2.1 (the superposition principle) implies that this function is a solution of (1) on I. From (7) we see that it satisfies the initial conditions y(x₀) = 0, y′(x₀) = 0. Now another solution of (1) satisfying the same initial conditions is y* ≡ 0. Since the coefficients p and q of (1) are continuous, Theorem 1 applies and gives uniqueness, that is, y ≡ y*, written out

Now since k₁ and k₂ are not both zero, this means linear dependence of y₁, y₂ on I.

(c) We prove the last statement of the theorem. If W(x₀) = 0 at an x₀ in I, we have linear dependence of y₁, y₂ on I by part (b), hence W ≡ 0 by part (a) of this proof. Hence in the case of linear dependence it cannot happen that W(x₁) ≠ 0 at an x₁ in I. If it does happen, it thus implies linear independence as claimed.

EXAMPLE 1 Illustration of Theorem 2

The functions y₁ = cos ωx and y₂ = sin ωx are solutions of y″ + ω²y = 0. Their Wronskian is

Theorem 2 shows that these solutions are linearly independent if and only if ω ≠ 0. Of course, we can see this directly from the quotient y₂/y₁. For ω = 0 we have y₂ = 0, which implies linear dependence (why?).

EXAMPLE 2 Illustration of Theorem 2 for a Double Root

A general solution of y″ − 2y′ + y = 0 on any interval is y = (c₁ + c₂x)e^x. (Verify!). The corresponding Wronskian is not 0, which shows linear independence of e^x and xe^x on any interval. Namely,

THEOREM 3 Existence of a General Solution

If p(x) and q(x) are continuous on an open interval I, then (1) has a general solution on I.

PROOF

By Theorem 1, the ODE (1) has a solution y₁(x) on I satisfying the initial conditions

and a solution y₂(x) on I satisfying the initial conditions

The Wronskian of these two solutions has at x = x₀ the value

Hence, by Theorem 2, these solutions are linearly independent on I. They form a basis of solutions of (1) on I, and y = c₁y₁ + c₂y₂ with arbitrary c₁, c₂ is a general solution of (1) on I, whose existence we wanted to prove.

THEOREM 4 A General Solution Includes All Solutions

If the ODE (1) has continuous coefficients p(x) and q(x) on some open interval I, then every solution y = Y(x) of (1) on I is of the form

where y₁, y₂ is any basis of solutions of (1) on I and C₁, C₂ are suitable constants.

Hence (1) does not have singular solutions(that is, solutions not obtainable from a general solution).

PROOF

Let y = Y(x) be any solution of (1) on I. Now, by Theorem 3 the ODE (1) has a general solution

on I. We have to find suitable values of c₁, c₂ such that y(x) = Y(x) on I. We choose any in I and show first that we can find values of c₁, c₂ such that we reach agreement at x₀, that is, y(x₀) and y′(x₀) = Y′(x₀). Written out in terms of (9), this becomes

We determine the unknowns c₁ and c₂. To eliminate c₂, we multiply (10a) by y′₂(x₀) and (10b) by −y₂(x₀) and add the resulting equations. This gives an equation for c₁. Then we multiply (10a) by −y′₁(x₀) and (10b) by y₁(x₀) and add the resulting equations. This gives an equation for c₂. These new equations are as follows, where we take the values of y₁, y′₁, y₂, y′₂, Y, Y′ at x₀.

Since y₁, y₂ is a basis, the Wronskian W in these equations is not 0, and we can solve for c₁ and c₂. We call the (unique) solution c₁ = C₁, c₂ = C₂. By substituting it into (9) we obtain from (9) the particular solution

Now since C₁, C₂ is a solution of (10), we see from (10) that

From the uniqueness stated in Theorem 1 this implies that y* and Y must be equal everywhere on I, and the proof is complete.

PROBLEM SET 2.6

1. Derive (6*) from (6).

2–8 BASIS OF SOLUTIONS. WRONSKIAN

Find the Wronskian. Show linear independence by using quotients and confirm it by Theorem 2.

• 2. e^4.0x, e^−1.5x
• 3. e^−0.4x, e^−2.6x
• 4. x, 1/x
• 5. x³, x²
• 6. e^−x cos ωx, e^−x sin ωx
• 7. cosh ax, sinh ax
• 8. x^k cos (ln x), x^k sin *ln x)

9–15 ODE FOR GIVEN BASIS. WRONSKIAN. IVP

(a) Find a second-order homogeneous linear ODE for which the given functions are solutions. (b) Show linear independence by the Wronskian. (c) Solve the initial value problem.

• 9. cos 5x, sin 5x, y(0) = 3, y′(0) = −5
• 10.
• 11. e^−2.5x cos 0.3x, e^−2.5x sin 0.3x, y(0) = 3, y′(0) = −7.5
• 12. x², x² ln x, y(1) = 4, y′(1) = 6
• 13. 1, e^−2x, y(0) = 1, y′(0) = −1
• 14. e^−kx cos πx, e^−kx sin πx, y(0) = 1, y′(0) = −k − π
• 15. cosh 1.8x, sinh 1.8x, y(0) = 14.20, y′(0) = 16.38
• 16. TEAM PROJECT. Consequences of the Present Theory. This concerns some noteworthy general properties of solutions. Assume that the coefficients p and q of the ODE (1) are continuous on some open interval I, to which the subsequent statements refer.

  (a) Solve y″ − y = 0 (a) by exponential functions, (b) by hyperbolic functions. How are the constants in the corresponding general solutions related?

  (b) Prove that the solutions of a basis cannot be 0 at the same point.

  (c) Prove that the solutions of a basis cannot have a maximum or minimum at the same point.

  (d) Why is it likely that formulas of the form (6*) should exist?

  (e) Sketch y₁(x) = x³ if x 0 and 0 if x < 0, y₂(x) = 0 if x 0 and x³ if x < 0. Show linear independence on −1 < x < 1. What is their Wronskian? What Euler–Cauchy equation do y₁, y₂ satisfy? Is there a contradiction to Theorem 2?

  (f) Prove Abel's formula⁶

  

  where c = W(y₁(x₀), y₂(x₀)). Apply it to Prob. 6. Hint: Write (1) for y₁ and for y₂. Eliminate q algebraically from these two ODEs, obtaining a first-order linear ODE. Solve it.

DEFINITION General Solution, Particular Solution

A general solution of the nonhomogeneous ODE (1) on an open interval I is a solution of the form

here, y_h = c₁y₁ + c₂y₂ is a general solution of the homogeneous ODE (2) on I and y_p is any solution of (1) on I containing no arbitrary constants.

A particular solution of (1) on I is a solution obtained from (3) by assigning specific values to the arbitrary constants c₁ and c₂ in y_h.

THEOREM 1 Relations of Solutions of (1) to Those of (2)

1. The sum of a solution y of (1) on some open interval I and a solution of (2) on I is a solution of (1) on I. In particular, (3) is a solution of (1) on I.
2. The difference of two solutions of (1) on I is a solution of (2) on I.

PROOF

1. Let L[y] denote the left side of (1). Then for any solutions y of (1) and of (2) on I,

   

2. For any solutions y and y* of (1) on I we have L[y − y*] = L[y] − L[y*] = r − r = 0.

Now for homogeneous ODEs (2) we know that general solutions include all solutions. We show that the same is true for nonhomogeneous ODEs (1).

THEOREM 2 A General Solution of a Nonhomogeneous ODE Includes All Solutions

If the coefficients p(x), q(x), and the function r(x) in (1) are continuous on some open interval I, then every solution of (1) on I is obtained by assigning suitable values to the arbitrary constants c₁ and c₂ in a general solution (3) of (1) on I.

PROOF

Let y* be any solution of (1) on I and x₀ any x in I. Let (3) be any general solution of (1) on I. This solution exists. Indeed, y_h = c₁y₁ + c₂y₂ exists by Theorem 3 in Sec. 2.6 because of the continuity assumption, and exists according to a construction to be shown in Sec. 2.10. Now, by Theorem 1(b) just proved, the difference Y = y* − y_p is a solution of (2) on I. At x₀ we have

Theorem 1 in Sec. 2.6 implies that for these conditions, as for any other initial conditions in I, there exists a unique particular solution of (2) obtained by assigning suitable values to c₁, c₂ in y_h. From this and y* = Y + y_p the statement follows.

EXAMPLE 1 Application of the Basic Rule (a)

Solve the initial value problem

Solution. Step 1. General solution of the homogeneous ODE. The ODE y″ + y = 0 has the general solution

Step 2. Solution y_p of the nonhomogeneous ODE. We first try y_p = Kx². Then y″_p = 2K. By substitution, 2K + Kx² = 0.001x². For this to hold for all x, the coefficient of each power of x(x² and x⁰) must be the same on both sides; thus K = 0.001 and 2K = 0, a contradiction.

The second line in Table 2.1 suggests the choice

Equating the coefficients of x², x, x⁰ on both sides, we have K₂ = 0.001, K₁ = 0, 2K₂ + K₀ = 0. Hence K₀ = −2K₂ = −0.002. This gives y_p = 0.001x² − 0.002, and

Step 3. Solution of the initial value problem. Setting x = 0 and using the first initial condition gives y(0) = A − 0.002 = 0, hence A = 0.002. By differentiation and from the second initial condition,

This gives the answer (Fig. 50)

Figure 50 shows y as well as the quadratic parabola y_p about which y is oscillating, practically like a sine curve since the cosine term is smaller by a factor of about 1/1000.

EXAMPLE 2 Application of the Modification Rule (b)

Solve the initial value problem

Solution. Step 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneous ODE is λ² + 3λ + 2.25 = (λ + 1.5)² = 0. Hence the homogeneous ODE has the general solution

Step 2. Solution of the nonhomogeneous ODE. The function e^−1.5x on the right would normally require the choice Ce^−1.5x. But we see from y_h that this function is a solution of the homogeneous ODE, which corresponds to a double root of the characteristic equation. Hence, according to the Modification Rule we have to multiply our choice function by x². That is, we choose

We substitute these expressions into the given ODE and omit the factor e^−1.5x. This yields

Comparing the coefficients of x², x, x⁰ gives 0 = 0, 0 = 0, 2C = −10, hence C = −5. This gives the solution y_p = −5x²e^−1.5x. Hence the given ODE has the general solution

Step 3. Solution of the initial value problem. Setting x = 0 in y and using the first initial condition, we obtain y(0) = c₁ = 1. Differentiation of y gives

From this and the second initial condition we have y′(0) = c₂ − 1.5c₁ = 0. Hence c₂ = 1.5c₁ = 1.5. This gives the answer (Fig. 51)

The curve begins with a horizontal tangent, crosses the x-axis at x = 0.6217 (where 1 + 1.5x − 5x² = 0) and approaches the axis from below as x increases.

EXAMPLE 3 Application of the Sum Rule (c)

Solve the initial value problem

Solution. Step 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneous ODE is

which gives the general solution y_h = c₁e^−x/2 + c₂e^−3x/2.

Step 2. Particular solution of the nonhomogeneous ODE. We write y_p = y_p1 + y_p2 and, following Table 2.1, (C) and (B),

Differentiation gives y′_p1 = −K sin x + M cos x, y″_p1 = −K cos x − M sin x and y′_p2 = 1, y″_p2 = 0. Substitution of y_p1 into the ODE in (7) gives, by comparing the cosine and sine terms,

hence K = 0 and M = 1. Substituting y_p2 into the ODE in (7) and comparing the x- and x⁰-terms gives

Hence a general solution of the ODE in (7) is

Step 3. Solution of the initial value problem. From y, y′ and the initial conditions we obtain

Hence c₁ = 3.1, c₂ = 0. This gives the solution of the IVP (Fig. 52)

PROBLEM SET 2.7

1–10 NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION

Find a (real) general solution. State which rule you are using. Show each step of your work.

1. y″ + 5y′ + 4y = 10e^−3x
2. 10y″ + 50y′ + 57.6y = cos x
3. y″ + 3y′ + 2y = 12x²
4. y″ − 9y = 18 cos πx
5. y″ + 4y′ + 4y = e^−x cos x
6. 
7. 
8. (3D² + 27I)y = 3 cos x + cos 3x
9. (D² − 16I)y = 9.6e^4x + 30e^x
10. (D² + 2D + I)y = 2x sin x

11–18 NONHOMOGENEOUS LINEAR ODEs: IVPs

Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.

• 11. y″ + 3y = 18x², y(0) = −3, y′(0) = 0
• 12. y″ + 4y = −12 sin 2x, y(0) = 1.8, y′(0) = 5.0
• 13. 8y″ − 6y′ + y = 6 cosh x, y(0) = 0.2, y′(0) = 0.05
• 14. y″ + 4y′ + 4y = e^−2x sin 2x, y(0) = 1, y′(0) = −1.5
• 15. (x²D² − 3xD + 3I)y = 3 ln x − 4, y(1) = 0, y′(1) = 1; y_p = ln x
• 16. (D² − 2D)y = 6e^2x − 4e^−2x, y(0) = −1, y′(0) = 6
• 17. (D² + 0.2D + 0.26I)y = 1.22e^0.5x, y(0) = 3.5, y′(0) = 0.35
• 18. (D² + 2D + 10I)y = 17 sin x − 37 sin 3x, y(0) = 6.6, y′(0) = −2.2
• 19. CAS PROJECT. Structure of Solutions of Initial Value Problems. Using the present method, find, graph, and discuss the solutions y of initial value problems of your own choice. Explore effects on solutions caused by changes of initial conditions. Graph y_p, y, y − y_p separately, to see the separate effects. Find a problem in which (a) the part of y resulting from y_h decreases to zero, (b) increases, (c) is not present in the answer y. Study a problem with y(0) = 0, y′(0) = 0. Consider a problem in which you need the Modification Rule (a) for a simple root, (b) for a double root. Make sure that your problems cover all three Cases I, II, III (see Sec. 2.2).
• 20. TEAM PROJECT. Extensions of the Method of Undetermined Coefficients. (a) Extend the method to products of the function in Table 2.1, (b) Extend the method to Euler–Cauchy equations. Comment on the practical significance of such extensions.

THEOREM 1 Steady-State Solution

After a sufficiently long time the output of a damped vibrating system under a purely sinusoidal driving force[see (2)]will practically be a harmonic oscillation whose frequency is that of the input.

PROBLEM SET 2.8

1. WRITING REPORT. Free and Forced Vibrations. Write a condensed report of 2–3 pages on the most important similarities and differences of free and forced vibrations, with examples of your own. No proofs.
2. Which of Probs. 1–18 in Sec. 2.7 (with x = time t) can be models of mass–spring systems with a harmonic oscillation as steady-state solution?

3–7 STEADY-STATE SOLUTIONS

Find the steady-state motion of the mass–spring system modeled by the ODE. Show the details of your work.

• 3. y″ + 6y′ + 8y = 42.5 cos 2t
• 4. y″ + 2.5y′ + 10y = −13.6 sin 4t
• 5. (D² + D + 4.25I)y = 22.1 cos 4.5t
• 6.
• 7. (4D² + 12D + 9I)y = 225 − 75 sin 3t

8–15 TRANSIENT SOLUTIONS

Find the transient motion of the mass–spring system modeled by the ODE. Show the details of your work.

• 8. 2y″ + 4y′ + 6.5y = 4 sin 1.5t
• 9. y″ + 3y′ + 3.25y = 3 cos t − 1.5 sin t
• 10. y″ + 16y = 56 cos 4t
• 11.
• 12. (D² + 2D + 5I)y = 4 cos t + 8 sin t
• 13. (D² + I)y = cos ωt, ω² ≠ 1
• 14. (D² + I)y = 5e^−t cos t
• 15. (D² + 4D + 8I)y = 2 cos 2t + sin 2t

16–20 INITIAL VALUE PROBLEMS

Find the motion of the mass–spring system modeled by the ODE and the initial conditions. Sketch or graph the solution curve. In addition, sketch or graph the curve of y − y_p to see when the system practically reaches the steady state.

• 16. y″ + 25y = 25 sin t, y(0) = 1, y′(0) = 1
• 17.
• 18. (D² + 8D + 17I)y = 474.5 sin 0.5t, y(0) = −5.4, y′(0) = 9.4
• 19.
• 20. (D² + 5I)y = cos πt − sin πt, y(0) = 0, y′(0) = 0
• 21. Beats. Derive the formula after (12) from (12). Can we have beats in a damped system?
• 22. Beats. Solve y″ + 25y = 99 cos 4.9t, y(0) = 2, y′(0) = 0. How does the graph of the solution change if you change (a) y(0), (b) the frequency of the driving force?
• 23. TEAM EXPERIMENT. Practical Resonance.

  (a) Derive, in detail, the crucial formula (16).

  (b) By considering dC*/dc show that C*(ω_max) increases as decreases.

  (c) Illustrate practical resonance with an ODE of your own in which you vary c, and sketch or graph corresponding curves as in Fig. 57.

  (d) Take your ODE with c fixed and an input of two terms, one with frequency close to the practical resonance frequency and the other not. Discuss and sketch or graph the output.

  (e) Give other applications (not in the book) in which resonance is important.

• 24. Gun barrel. Solve y″ + y = 1 − t²/π² if 0 t π and 0 if t → ∞; here, y(0) = 0, y′(0) = 0. This models an undamped system on which a force F acts during some interval of time (see Fig. 59), for instance, the force on a gun barrel when a shell is fired, the barrel being braked by heavy springs (and then damped by a dashpot, which we disregard for simplicity). Hint: At π both y and y′ must be continuous.

  

  Fig. 59. Problem 24

• 25. CAS EXPERIMENT. Undamped Vibrations.

  (a) Solve the initial value problem y″ + y = cos ωt, ω² ≠ 1, y(0) = 0, y′(0) = 0. Show that the solution can be written

  

  (b) Experiment with the solution by changing ω to see the change of the curves from those for small ω (>0) to beats, to resonance, and to large values of (see Fig. 60).

  

  Fig. 60. Typical solution curves in CAS Experiment 25

EXAMPLE 1 RLC-Circuit

Find the current I(t) in an RLC-circuit with R = 11 Ω (ohms), L = 0.1 H (henry), C = 10⁻²F (farad), which is connected to a source of EMF E(t) = 110 sin (60 · 2πt) = 110 sin 377 t (hence 60 Hz = 60 cycles/sec, the usual in the U.S. and Canada; in Europe it would be 220 V and 50 Hz). Assume that current and capacitor charge are 0 when t = 0.

Solution. Step 1. General solution of the homogeneous ODE. Substituting R, L, C and the derivative E′(t) into (1), we obtain

Hence the homogeneous ODE is 0.1I″ + 11I′ + 100I = 0. Its characteristic equation is

The roots are λ₁ = −10 and λ₂ = −100. The corresponding general solution of the homogeneous ODE is

Step 2. Particular solution I_p of (1). We calculate the reactance S = 37.7 − 0.3 = 37.4 and the steady-state current

with coefficients obtained from (4) (and rounded)

Hence in our present case, a general solution of the nonhomogeneous ODE (1) is

Step 3. Particular solution satisfying the initial conditions. How to use Q(0) = 0? We finally determine c₁ and c₂ from the in initial conditions I(0) = 0 and Q(0) = 0. From the first condition and (6) we have

We turn to Q(0) = 0. The integral in (1′) equals ∫I dt = Q(t); see near the beginning of this section. Hence for t = 0, Eq. (1′) becomes

Differentiating (6) and setting t = 0, we thus obtain

The solution of this and (7) is c₁ = −0.323, c₂ = 3.033. Hence the answer is

You may get slightly different values depending on the rounding. Figure 63 shows I(t) as well as I_p(t), which practically coincide, except for a very short time near t = 0 because the exponential terms go to zero very rapidly. Thus after a very short time the current will practically execute harmonic oscillations of the input frequency 60 Hz = 60 cycles/sec. Its maximum amplitude and phase lag can be seen from (5), which here takes the form

PROBLEM SET 2.9

1–6 RLC-CIRCUITS: SPECIAL CASES

1. RC-Circuit. Model the RC-circuit in Fig. 64. Find the current due to a constant E.

   

   Fig. 64. RC-circuit

   

   Fig. 65. Current 1 in Problem 1

2. RC-Circuit. Solve Prob. 1 when E = E₀ sin ωt and R, C, E₀, and ω are arbitrary.
3. RL-Circuit. Model the RL-circuit in Fig. 66. Find a general solution when R, L, E are any constants. Graph or sketch solutions when L = 0.25 H, R = 10 Ω, and E = 48 V.

   

   Fig. 66. RL-circuit

   

   Fig. 67. Currents in Problem 3

4. RL-Circuit. Solve Prob. 3 when E = E₀ sin ωt and R, L, E₀, and are arbitary. Sketch a typical solution.

   

   Fig. 68. Typical current in Problem 4

5. LC-Circuit. This is an RLC-circuit with negligibly small R (analog of an undamped mass–spring system). Find the current when L = 0.5 H, C = 0.005 F, and E = sin t V, assuming zero initial current and charge.

   

   Fig. 69. LC-circuit

6. LC-Circuit. Find the current when L = 0.5 H, C = 0.005 F, E = 2t² V, and initial current and charge zero.

7–18 GENERAL RLC-CIRCUITS

• 7. Tuning. In tuning a stereo system to a radio station, we adjust the tuning control (turn a knob) that changes C (or perhaps L) in an RLC-circuit so that the amplitude of the steady-state current (5) becomes maximum. For what C will this happen?

8–14 Find the steady-state current in the RLC-circuit in Fig. 61 for the given data. Show the details of your work.

• 8. R = 4 Ω, L = 0.5 H, C = 0.1 F, E = 500 sin 2t V
• 9. R = 4 Ω L = 0.1 H, C = 0.05 F, E = 110 V
• 10. R = 2 Ω, L = 1 H, E = 157 sin 3t V
• 11.
• 12. R = 0.2 Ω, L = 0.1 H, C = 2 F, E = 220 sin 314t V
• 13.
• 14. Prove the claim in the text that if R ≠ 0 (hence R > 0), then the transient current approaches I_p as t → ∞.
• 15. Cases of damping. What are the conditions for an RLC-circuit to be (I) overdamped, (II) critically damped, (III) underdamped? What is the critical resistance R_crit (the analog of the critical damping constant )?

16–18 Solve the initial value problem for the RLC-circuit in Fig. 61 with the given data, assuming zero initial current and charge. Graph or sketch the solution. Show the details of your work.

• 16. R = 8 Ω, L = 0.2 H, C = 12.5 · 10⁻³F, E = 100 sin 10t V
• 17. R = 6 Ω, L = 1 H, C = 0.04 F, E = 600 (cos t + 4 sin t) V
• 18. R = 18 Ω, L = 1 H, C = 12.5 · 10⁻³F, E = 820 cos 10t V
• 19. WRITING REPORT. Mechanic-Electric Analogy. Explain Table 2.2 in a 1–2 page report with examples, e.g., the analog (with L = 1 H) of a mass–spring system of mass 5 kg, damping constant 10 kg sec, spring constant 60 kg/sec², and driving force 220 cos 10t kg/sec.
• 20. Complex Solution Method. Solve , by substituting I_p = Ke^iωt (K unknown) and its derivatives and taking the real part I_p of the solution . Show agreement with (2), (4). Hint: Use (11) e^iωt = cos ωt + i sin ωt; cf. Sec. 2.2, and i² = −1.

EXAMPLE 1 Method of Variation of Parameters

Solve the nonhomogeneous ODE

Solution. A basis of solutions of the homogeneous ODE on any interval is y₁ = cos x, y₂ = sin x. This gives the Wronskian

From (2), choosing zero constants of integration, we get the particular solution of the given ODE

Figure 70 shows y_p and its first term, which is small, so that x sin x essentially determines the shape of the curve of y_p. (Recall from Sec. 2.8 that we have seen x sin x in connection with resonance, except for notation.) From y_p and the general solution y_h = c₁y₁ + c₂y₂ of the homogeneous ODE we obtain the answer

Had we included integration constants −c₁, c₂ in (2), then (2) would have given the additional c₁ cos x + c₂ sin x = c₁y₁ + c₂y₂ that is, a general solution of the given ODE directly from (2). This will always be the case.

Fig. 70. Particular solution y_p and its first term in Example 1

PROBLEM SET 2.10

1–13 GENERAL SOLUTION

Solve the given nonhomogeneous linear ODE by variation of parameters or undetermined coefficients. Show the details of your work.

1. y″ + 9y = sec 3x
2. y″ + 9y = csc 3x
3. x²y″ − 2xy′ + 2y = x³ sin x
4. y″ − 4y′ + 5y = e^2x csc x
5. y″ + y = cos x − sin x
6. (D² + 6D + 9I)y = 16e^−3x/(x² + 1)
7. (D² − 4D + 4I)y = 6e^2x/x⁴
8. (D² + 4I)y = cosh 2x
9. (D² − 2D + I)y = 35x^3/2e^x
10. (D² + 2D + 2I)y = 4e^−x sec³x
11. (x²D² − 4xD + 6I)y = 21x⁻⁴
12. (D² − I)y = 1/cosh x
13. (x²D² + xD − 9I)y = 38x⁵
14. TEAM PROJECT. Comparison of Methods. Invention. The undetermined-coefficient method should be used whenever possible because it is simpler. Compare it with the present method as follows.

    (a) Solve y″ + 4y′ + 3y = 65 cos 2x by both methods, showing all details, and compare.

    (b) Solve y″ − 2y′ + y = r₁ + r₂, r₁ = 35x^3/2e^xr₂ = x² by applying each method to a suitable function on the right.

    (c) Experiment to invent an undetermined-coefficient method for nonhomogeneous Euler–Cauchy equations.

Second-order linear ODEs are particularly important in applications, for instance, in mechanics (Secs. 2.4, 2.8) and electrical engineering (Sec. 2.9). A second-order ODE is called linear if it can be written

(If the first term is, say, f(x)y″ divide by f(x) to get the “standard form” (1) with y″ as the first term.) Equation (1) is called homogeneous if r(x) is zero for all x considered, usually in some open interval; this is written r(x) ≡ 0. Then

Equation (1) is called nonhomogeneous if r(x) 0 (meaning r(x) is not zero for some x considered).

For the homogeneous ODE (2) we have the important superposition principle (Sec. 2.1) that a linear combination y = ky₁ + ly₂ of two solutions y₁, y₂ is again a solution.

Two linearly independent solutions y₁, y₂ of (2) on an open interval I form a basis (or fundamental system) of solutions on I. and y = c₁y₁ + c₂y₂ with arbitrary constants c₁, c₂ a general solution of (2) on I. From it we obtain a particular solution if we specify numeric values (numbers) for c₁ and c₂ usually by prescribing two initial conditions

(2) and (3) together form an initial value problem. Similarly for (1) and (3).

For a nonhomogeneous ODE (1) a general solution is of the form

Here y_h is a general solution of (2) and y_p is a particular solution of (1). Such a y_p can be determined by a general method (variation of parameters, Sec. 2.10) or in many practical cases by the method of undetermined coefficients. The latter applies when (1) has constant coefficients p and q, and r(x) is a power of x, sine, cosine, etc. (Sec. 2.7). Then we write (1) as

The corresponding homogeneous ODE y′ + ay′ + by = 0 has solutions y = e^λx, where λ is a root of

Hence there are three cases (Sec. 2.2):

Here ω* is used since ω is needed in driving forces.

Important applications of (5) in mechanical and electrical engineering in connection with vibrations and resonance are discussed in Secs. 2.4, 2.7, and 2.8.

Another large class of ODEs solvable “algebraically” consists of the Euler–Cauchy equations

These have solutions of the form y = x^m where m is a solution of the auxiliary equation

Existence and uniqueness of solutions of (1) and (2) is discussed in Secs. 2.6 and 2.7, and reduction of order in Sec. 2.1.